# At the equator a 1000 turn coil with a cross sectional area of 300 cm2 and a resistance of 15 #Omega# is aligned with its plane perpendicular to the earth’s magnetic field of 7*10^-5 T . If the coil is flipped over how much charge flows through it?

##### 1 Answer

Important information missing: Time taken to flip over the coil

Let this be

We know that the average induced current

#I_(av)=varepsilon/R# .........(1)

where#varepsilon# is the induced emf produced due to change in flux linked with the coil and#R# is the resistance of coil.

We also know from Lenz's law that the induced emf in the coil is the rate at which the magnetic flux linked with the coil changes and opposes it.

#varepsilon=-(Deltaphi)/t# .......(2)

The coil of

#Deltaphi=-2phi=-2NBA# ......(3)

Rewriting (1) with the helps of (2) and (3) we get

#I_(av)=(2NBA)/(Rt)# ......(4)

Now charge through coil in time

#I=(Q)/t#

#=>Q=It# ......(5)

Using (4) we get

#Q=(2NBA)/(Rt)xxt#

#=>Q=(2NBA)/(R)#

Inserting given values in SI units we get

#Q=(2xx1000xx(7xx10^-5)xx300/10^4)/15#

#=>Q=2.8xx10^-4\ C#